Question

10 баллов. решить неравенство.
log1/2(2x+4)>log1/2(x+1);

Answer 1

$log_{_frac{1}{2}}(2x+4) textgreater log_{_frac{1}{2}}(x+1) ODZ: left { {{2x+4 textgreater 0} atop {x+1 textgreater 0}} right. left { {{x textgreater -2} atop {x textgreater -1}} right. x textgreater -1 \f(t)=log_{_frac{1}{2}}t - spadayushaya funkziya, poetomu: \2x+4 textless x+1 \x textless -3 - ne vhodit v ODZ \OTBET: neravenstvo ne imeet resheniy$